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% \subsection*{1.9. French}

\textbf{Lemme 1.9.3}

Soient deux systèmes $(V_0, \tau_0, e_0), (V_1, \tau_1, e_1)$ comme en 1.9. 

On a
\[
\left\vert \sup_{j \leq i} (-v\nabla_{\tau_i}^j e_1) - \sup_{j \leq i} (-v\nabla_{\tau_0}^j e_0)\right\vert \leq c^{te}.
\]

Il suffit d'établir une inégalité (1.9.3) lorsqu'on change une seule des données $V_0$, $\tau_0$, $e$. 

Le cas où on change seulement le réseau de référence $V_0$ résulte de (1.8.1).

On utilisera systématiquement que, pour $f \in K$, on a par 1.4.3 (notation de 1.2.3)
\begin{equation}
v(\partial_{\tau_1} f) \geq v(f). \tag{1.9.4}
\end{equation}

Si $e$ et $f$ sont deux bases, on a $e = fa$ avec $a \in \mathrm{GL}_n(K)$, d'où
\[
\nabla_\tau^i(e) = \sum_j \binom{i}{j} \nabla_\tau^j(f)\cdot \nabla_\tau^{i-j} a,
\]
et, par (1.9.4),
\[
v(\nabla_\tau^i(e)) \geq \inf_{j \leq i} v(\nabla_\tau^j f) + c^{te},
\]
d'où
\[
\sup_{j \leq i} (-v\nabla_\tau^j e) - \sup_{j \leq i} (-v\nabla_\tau^j f) \leq c^{te}.
\]

Renversant les rôles de $e$ et $f$, on a de même
\[
\sup_{j \leq i} (-v\nabla_\tau^j f) - \sup_{j \leq i} (-v\nabla_\tau^j e) \leq c^{te},
\]
d'où l'estimation 1.9.3 pour un changement de base.


%\page*{- 47 -}

Si $\tau$ et $\sigma$ sont deux vecteurs comme en 1.9, on a $\sigma = f\tau$ avec $f$ inversible, d'où
\[
\nabla_\sigma = f \nabla_\tau \quad (f \in \mathcal{O}^*)
\]
et on vérifie par récurrence que
\[
\nabla_\sigma^i e = \sum_{j \leq i} \varphi_j \nabla_\tau^j e \quad (\varphi_j \in \mathcal{O}).
\]

On en déduit que
\[
v(\nabla_\sigma^i e) \geq \inf v(\nabla_\tau^j e),
\]
d'où
\[
\sup_{j \leq i} (-v\nabla_\sigma^j e) \leq \sup_{j \leq i} (-v\nabla_\tau^j e).
\]

Renversant les rôles de $\sigma$ et $\tau$, on conclut que
\begin{equation}
\sup_{j \leq i} (-v\nabla_\sigma^j e) = \sup_{j \leq i} (-v\nabla_\tau^j e). \tag{1.9.5}
\end{equation}

D'après 1.9.1 et 1.9.3, il suffit, pour prouver 1.9, de prouver une majoration de type (1.9.2) pour un choix de $(V_0, \tau, e)$.


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\newpage 

% \subsection*{1.9. English}

\textbf{Lemma 1.9.3}

Let $(V_0, \tau_0, e_0)$ and $(V_1, \tau_1, e_1)$ be two such triples as in 1.9. Then
\[
\left\vert \sup_{j \leq i} (-v\nabla_{\tau_1}^j e_1) - \sup_{j \leq i} (-v\nabla_{\tau_0}^j e_0)\right\vert \leq c^{te}.
\]

It suffices to establish inequality (1.9.3) when only one of the data $V_0$, $\tau_0$, or $e$ is changed.

The case where only the reference lattice $V_0$ is changed follows from (1.8.1).

We will systematically use the fact that, for $f \in K$, we have by (1.4.3) (with 1.2.3)%(with the notation of 1.2.3)
\begin{equation}
v(\partial_{\tau_1} f) \geq v(f). \tag{1.9.4}
\end{equation}

If $e$ and $f$ are two bases, then $e = f a$ with $a \in \mathrm{GL}_n(K)$, so
\[
\nabla_\tau^i(e) = \sum_j \binom{i}{j} \nabla_\tau^j(f)\cdot \nabla_\tau^{i-j} a,
\]
and by (1.9.4),
\[
v(\nabla_\tau^i(e)) \geq \inf_{j \leq i} v(\nabla_\tau^j f) + c^{te},
\]
hence
\[
\sup_{j \leq i} (-v\nabla_\tau^j e) - \sup_{j \leq i} (-v\nabla_\tau^j f) \leq c^{te}.
\]

Interchanging the roles of $e$ and $f$, we similarly obtain
\[
\sup_{j \leq i} (-v\nabla_\tau^j f) - \sup_{j \leq i} (-v\nabla_\tau^j e) \leq c^{te},
\]
which yields estimate (1.9.3) under a change of basis.

%\page{- 47 -}

If $\tau$ and $\sigma$ are two vectors as in 1.9, then $\sigma = f \tau$ with $f$ invertible, so
\[
\nabla_\sigma = f \nabla_\tau \quad (f \in \mathcal{O}^),
\]
and one checks inductively that
\[
\nabla_\sigma^i e = \sum_{j \leq i} \varphi_j \nabla_\tau^j e \quad (\varphi_j \in \mathcal{O}).
\]

It follows that
\[
v(\nabla_\sigma^i e) \geq \inf_{j \leq i} v(\nabla_\tau^j e),
\]
hence
\[
\sup_{j \leq i} (-v\nabla_\sigma^j e) \leq \sup_{j \leq i} (-v\nabla_\tau^j e).
\]

Reversing the roles of $\sigma$ and $\tau$, we conclude that
\begin{equation}
\sup_{j \leq i} (-v\nabla_\sigma^j e) = \sup_{j \leq i} (-v\nabla_\tau^j e). \tag{1.9.5}
\end{equation}

By Lemmas 1.9.1 and 1.9.3, it suffices, in order to prove Theorem 1.9, to establish an estimate of type (1.9.2) for a single convenient choice of $(V_0, \tau, e)$.

